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Ionic Equilibrium | Chemistry - Buffer Solution | 12th Chemistry : UNIT 8 : Ionic Equilibrium

Chapter: 12th Chemistry : UNIT 8 : Ionic Equilibrium

Buffer Solution

Do you know that our blood maintains a constant pH, irrespective of a number of cellular acid – base reactions. Is it possible to maintain a constant hydronium ion concentration in such reactions? Yes, it is possible due to buffer action.

Buffer Solution

Do you know that our blood maintains a constant pH, irrespective of a number of cellular acid – base reactions. Is it possible to maintain a constant hydronium ion concentration in such reactions? Yes, it is possible due to buffer action.

Buffer is a solution which consists of a mixture of a weak acid and its conjugate base (or) a weak base and its conjugate acid. This buffer solution resists drastic changes in its pH upon addition of a small quantities of acids (or) bases, and this ability is called buffer action. The buffer containing carbonic acid (H2 CO3 ) and its conjugate base HCO-3 is present in our blood. There are two types of buffer solutions.

1. Acidic buffer solution : a solution containing a weak acid and its salt.

Example : solution containing acetic acid and sodium acetate

2. Basic buffer solution : a solution containing a weak base and its salt.

Example : Solution containing NH4OH and NH4Cl

 

Buffer action

To resist changes in its pH on the addition of an acid (or) a base, the buffer solution should contain both acidic as well as basic components so as to neutralize the effect of added acid (or) base and at the same time, these components should not consume each other.

Let us explain the buffer action in a solution containing CH3COOH and CH3COONa.

The dissociation of the buffer components occurs as below.

CH3COOH (aq) CH3 – COO-(aq) + H3O+(aq)


If an acid is added to this mixture, it will be consumed by the conjugate base CH3COO- to form the undissociated weak acid i.e, the increase in the concentration of H+ does not reduce the pH significantly.

CH3COO - (aq) + H+ (aq) CH3COOH (aq)

If a base is added, it will be neutralized by H3O+, and the acetic acid is dissociated to maintain the equlibrium. Hence the pH is not significantly altered.


These neutralization reactions are identical to those reactions that we have already discussed in common ion effect.

Let us analyse the effect of the addition of 0.01 mol of solid sodium hydroxide to one litre of a buffer solution containing 0.8 M CH3COOH and 0.8 M CH3COONa . Assume that the volume change due to the addition of NaOH is negligible. (Given: Ka for CH3COOH is 1.8 ×10-5 )


The dissociation constant for CH3COOH is given by

Ka =[CH3COO-][H+] / [CH3COOH]

[H+ ]=Ka [CH3COOH ] / [CH3COO- ]

The above expression shows that the concentration of H+ is directly proportional to

[CH3COOH] / [CH3COO-] .

Let the degree of dissociation of CH3COOH be α then,

[CH3COOH] = 0.8-α and [CH3COO- ] = α+0.8

[H+] = Ka [(0.8-α)/(0.8+α)]

α<<0.8,     0.8-α 0.8 and 0.8+α 0.8

[H+ ] = [Ka (0.8)] / (0.8)

 ⇒ [H+ ]= Ka

Given that

Ka for CH3COOH is 1.8 ×10-5

[H+] = 1.8 ×105 ; pH = - log (1.8 ×10-5 )

 =5 - log 1.8

= 5 - 0.26

pH = 4.74

Calculation of pH after adding 0.01 mol NaOH to 1 litre of buffer.

Given that the volume change due to the addition of NaOH is negligible [OH-] = 0.01M .

The consumption of OH- are expressed by the following equations.

CH3 COOH (aq) CH3COO-(aq) + H+ (aq)

0.8-α                          α                                      α

CH3 COONa(aq) CH3COO- (aq) + Na+ (aq)

             0. 8                       0.8                       0.8

CH3COOH + OH- (aq) CH3COO- (aq) + H2O (l)

[CH3COOH] = 0.8 - α - 0.01 = 0.79 – α

[CH3COO-] = α+0.8+0.01= 0.81+α     α<<0.8;

0.79 - α = 0.79 and 0.81 + α = 0.81

[H+ ] = (1.8 ×105 ) × (0.79/0.81)

[H+ ] = 1.76 ×105

pH = - log ( 1.76 ×10-5 )

=5 - log 1.76

=5 - 0.25

pH = 4.75

The addition of a strong base (0.01 M NaOH) increased the pH only slightly ie., from 4.74 to 4.75 . So, the buffer action is verified.


Evaluate yourself – 8

a. Explain the buffer action in a basic buffer containing equimolar ammonium hydroxide and ammonium chloride.

b. Calculate the pH of a buffer solution consisting of 0.4M CH3COOH and 0.4M CH3COONa . What is the change in the pH after adding 0.01 mol of HCl to 500ml of the above buffer solution. Assume that the addition of HCl causes negligible change in the volume. Given: ( Ka = 1 . 8 ×105. )

a) Answer

Dissociation of buffer components

NH4OH (aq) ↔ NH4+ (aq) + OH- (aq)

NH4 Cl NH4+ +Cl+

Addition of H+

The added H+ ions are neutralized by NH4OH and there is no appreciable decrease in pH.

NH4OH(aq) + H+ → NH4+ (aq) + H2O(l)

Addition of OH–

NH4 + (aq) + OH- (aq) NH4OH (aq)

The added OH- ions react with NH4+ to produce unionized NH4OH. Since NH4OH is a weak base, there is no appreciable increase in pH

b) Answer

pH of buffer

CH3COOH(aq) ↔ CH3COO - (aq) + H + (aq)

0.4-α                   ↔            α         +         α

CH3 COONa(aq) CH 3 COO - (aq) + Na+ (aq)

0.4       →             0.4        +         0. 4

[H+ ]= K a [CH 3 COOH] / [CH3COO- ]

[CH3COOH] = 0.4 - α ≈ 0.4

[CH3COO- ] = 0.4 + α ≈ 0.4

[H+] = Ka (0.4) / (0.4)

[H+] = 1.8 ×10-5

pH = - log (1.8 ×10-5 ) = 4.74

Addition of 0.01 mol HCl to 500ml of buffer

Added [H+ ] =  0.01 mol / 500 mL = 0.01 mol / 1/2L

 = 0.02M

CH3 COOH(aq) ↔ CH3COO- (aq) + H+ (aq)

0.4-α ↔ α  + α

CH3 COONa CH3COO- +Na+

0.4 0.4 + 0.4

CH3COO- + HCl CH3COOH+Cl-

(0.02) + 0.02 0.02 + 0.02

[CH3COOH] = 0.4 - α + 0.02 = 0.42 - α ≈ 0.42

[CH3COO- ] = 0.4 + α - 0.02 = 0.38 + α ≈ 0.38

[H+] = [(1.8 × 10-5 ) (0.42)] / (0.38)

[H+ ] = 1.99 × 10-5

pH = - log (1.99 ×10-5 )

 = 5 - log 1.99

 = 5 - 0.30

 = 4.70

 

Buffer capacity and buffer index

The buffering ability of a solution can be measured in terms of buffer capacity. Vanslyke introduced a quantity called buffer index, β, as a quantitative measure of the buffer capacity. It is defined as the number of gram equivalents of acid or base added to 1 litre of the buffer solution to change its pH by unity.


Here,

dB = number of gram equivalents of acid / base added to one litre of buffer solution.

d(pH) = The change in the pH after the addition of acid / base.

 

Henderson – Hasselbalch equation

We have already learnt that the concentration of hydronium ion in an acidic buffer solution depends on the ratio of the concentration of the weak acid to the concentration of its conjugate base present in the solution i.e.,


The weak acid is dissociated only to a small extent. Moreover, due to common ion effect, the dissociation is further suppressed and hence the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid. Similarly, the concentration of the conjugate base is nearly equal to the initial concentration of the added salt.


Here [acid] and [salt] represent the initial concentration of the acid and salt, respectively used to prepare the buffer solution

Taking logarithm on both sides of the equation


 

Example 8.6

Find the pH of a buffer solution containing 0.20 mole per litre sodium acetate and 0.18 mole per litre acetic acid. Ka for acetic acid is 1.8 × 10-5 .

pH = pKa + log {[salt]/[acid]}

Given that Ka = 1.8 × 10-5

pKa = −log(1.8 ×105 ) = 5 log1.8

= 5 -0.26

= 4.74

pH = 4.74 + log (0.20/0.18)

= 4.74 + log (10/9) = 4.74 + log 10 - log 9

= 4.74 + 1 - 0.95 = 5.74 - 0.95

= 4.79

 

Example 8.7

What is the pH of an aqueous solution obtained by mixing 6 gram of acetic acid and 8.2 gram of sodium acetate making the volume equal to 500 ml. (Given: Ka for acetic acid is 1.8 × 10-5 )

According to Henderson – Hasselbalch equation,

pH = pKa + log {[salt]/[acid]}

PKa =-logKa =-log(1.8 ×10-5 )= 4.74 (Refer previous example)

Salt = Number of moles of sodium acetate / Volume of the solution ( litre )

Number of moles of sodium acetate = mass of sodium acetate / molar mass of sodium acetate


pH = 4.74 + log 1

pH = 4.74 + 0 = 4.74

 

Evaluate yourself – 9

a. How can you prepare a buffer solution of pH 9. You are provided with 0.1M NH4OH solution and ammonium chloride crystals. (Given: pKb for NH4OH is 4.7 at 25 C .

b. What volume of 0.6M sodium formate solution is required to prepare a buffer solution of pH 4.0 by mixing it with 100ml of 0.8M formic acid. (Given: pKa for formic acid is 3.75. )

a) answer

 pOH = pKb + log ([salt]/[base])

We know that

pH + pOH = 14

9 + pOH = 14

pOH = 14 - 9 = 5

 5 = 4.7 + log ([NH4Cl]/[NH4OH])

0.3=log ( [NH4Cl]/0.1)

[NH4Cl] / 0.1  = antilog of (0.3)

[NH4Cl] = 0.1M × 1.995

 = 0.1995 M

 = 0.2M

Amount of NH4Cl required to prepare 1 litre 0.2M solution

 = Strength of NH4Cl × molar mass of NH4Cl

 = 0.2 × 53.5

 = 10.70 g

10.70 g ammonium chloride is dissolved in water and the solution is made up to one litre to get 0.2M solution. On mixing equal volume of the given NH4OH solution and the prepared NH4Cl solution will give a buffer solution with required pH value (pH = 9).

b) answer

pH = pKa +log ( [salt]/[acid] )

 4 = 3.75+log ([sodium formate] /[formic acid])

[Sodium formate] = number of moles of HCOONa = 0.6 × V × 10−3

[formic acid] = number of moles of HCOOH

 = 0.8 × 100 × 10−3

= 80 × 10−3

4 = 3.75 + log (0.6V/80)

0.25 = log (0.6V/80)

antilog of 0.25 = 0.6V/80

0.6V = 1.778 × 80

 = 1.78 × 80

 = 142.4

V = 142.4 mL / 0.6 = 237.33mL

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